Question

How do you differentiate `y=tanx/(2x^3)`?

Answer 1

`dy/dx = ( x^2sec^2x - 3tanx ) / (2x^4)`

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

`d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`, or less formally, `(u/v)' = (v(du)-u(dv))/v^2`

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with `y=(tanx)/(2x^3)` Then

`{ ("Let "u=tanx, => , (du)/dx=sec^2x), ("And "v=2x^3, =>, (dv)/dx=6x^2 ) :}`

`:. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`
`:. " " dy/dx = ( (2x^3)(sec^2x) - (tanx)(6x^2) ) / (2x^3)^2`
`:. " " dy/dx = ( 2x^3sec^2x - 6x^2tanx ) / (4x^6)`
`:. " " dy/dx = ( x^2sec^2x - 3tanx ) / (2x^4)`